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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

Consider the equation, 4x^2 + 9y^2 - 64 + 72y + 364 = 0. Tell what conic section this represents and sketch its graph.

This is what I did:

(4x^2 - 64x. . ..) + (9y^2 + 72. . . .) = -364

4(x^2 - 16x. . ..) + 9(y^2 + 8y. . . .) = -364

4(x^2 - 16x + 1) + 9(y^2 + 8y + 8 ) = -364 + __

4(x^2 - 16x + 1) + 9(y^2 + 8y + 8 ) = -364 + 4(1) + 9(8)

4(x^2 - 16x + 1) + 9(y^2 + 8y + 8 ) = -364 + 4 + 72

4(x - 1)^2 + 9(y + 0.9)^2 = -288

[4(x - 1)^2] / -288 + [9(y + 0.9)^2] / -288 = -288 / -288

or

(x - 1)^2 / (sqrt(-72))^2 + (y + 0.9)^2 / ((sqrt(-32)) = 1

Is this correct?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I'm not sure what you've done here. You've started well, by separating the x terms from the y terms, and then it looks like you've tried to complete the square, but the steps don't look right.

(I'm assuming that where you've written 64, you mean 64x)

4x² + 9y² - 64x + 72y + 364 = 0

4(x² - 16x) + 9(y² + 8y) + 364 = 0

4(x² - 16x + 64) - 256 + 9(y² + 8y + 16) - 144 + 364 = 0

4(x-8)² + 9(y+4)² = 36

[(x-8)²]/9 + [(y+4)²]/4 = 1

That's the canonical form of an ellipse, and so we know what the conic section now is.

To draw it, you'd need to work out how to transform the standard ellipse to form this one.

Why did the vector cross the road?

It wanted to be normal.

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